 Solution

For an admissible Laplace integral solution, differentiation under the integral sign and integration by parts giveChoose endpoints for which the boundary term vanishes. Setting the density in braces to zero givesTwo suitable contours are the rays ${\Gamma }_{+}:0\to +i\infty$ and ${\Gamma }_{-}:0\to -i\infty$, with a continuous square-root branch on each ray. The Gaussian decays at infinity, and $\zeta g=O({\zeta }^{1/2})$ tends to zero at the finite endpoint. The integrals converge for every finite $z$.

To check independence rather than merely exhibit different paths, defineWith unit normalization, the two solutions satisfy $w_{\pm }(0)=e^{\pm i\pi /4}{J}_0$ and $w_{\pm }^{\prime }(0)=\pm i{e}^{\pm i\pi /4}{J}_1$. Their Wronskian at zero is $-2i{J}_0{J}_1\ne 0$, so they form an independent pair.

Strictly, the printed necessity claim requires an injective contour-transform class as well as vanishing endpoint terms. An arbitrary contour representation may contain a null weight in a contour Laplace representation, so a nonzero solution alone does not force every representing $g$ to have the displayed form. The calculation constructs the canonical density and two valid independent solutions; it does not assume uniqueness for unrestricted contour densities.